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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: xshadow on January 24, 2017, 02:03:55 PM

Title: oxidation number of SF5 -
Post by: xshadow on January 24, 2017, 02:03:55 PM
HI, I don't understand what is the oxidation number for SF5- and his meaning in this "complex"


The lewis structure of SF5 - is  a central sulfur atom with 5 single bond with F atoms (or F- ions) and a lone pair on S.

a)Using the formula ,SF5- I should have:   -1* 5(F atoms) + n.o S =  -1
So  n.o. S = +4      (verify: 4*1  -1*5 = -1 ...correct)

b) But if I see the structure and I apply the definition of Oxidation number, i.e. :" is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component"
I have 5 bonds with F (more electronegative)...so should be +5


Why there is this difference?
Thanks
Title: Re: oxidation number of SF5 -
Post by: AWK on January 24, 2017, 02:16:53 PM
But in the case of ion - sum of oxidation numbers is equal to the charge of ion (this is just math).
Title: Re: oxidation number of SF5 -
Post by: xshadow on January 24, 2017, 02:36:14 PM
Quote from: AWK on January 24, 2017, 02:16:53 PM
But in the case of ion - sum of oxidation numbers is equal to the charge of ion (this is just math).

so the oxidation state of sulfur is +4

So the second definition, b)  , can't work  because it gives +5

Thanks
Title: Re: oxidation number of SF5 -
Post by: sjb on January 24, 2017, 04:17:42 PM
Quote from: xshadow on January 24, 2017, 02:36:14 PM
so the oxidation state of sulfur is +4

So the second definition, b)  , can't work  because it gives +5

Thanks

If SF5- were completely ionic, you can consider the formalism SF51-  :rarrow: SF4 + F-.

Where is the contradiction in the oxidation state of sulfur now?